NCERT Solutions for Exercise 8.4 Class 10 Maths Chapter 8 - Introduction to Trigonometry

NCERT Solutions for Exercise 8.4 Class 10 Maths Chapter 8 - Introduction to Trigonometry

Updated on 30 Apr 2025, 03:22 PM IST

Trigonometric identities provide a foundation that enables quick solutions to multiple mathematical expressions. This exercise demonstrates applying basic trigonometric ratios of sine, cosine and tangent when performing standard identity simplifications and verifications. The linkage among different trigonometric functions becomes possible through these identities, which enable us to show complex expressions in separate steps. Learning trigonometric connections helps students develop their logical reasoning skills, leading to advanced trigonometric abilities needed for engineering, architecture and physical science fields.

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  1. NCERT Solutions Class 10 Maths Chapter 8: Exercise 8.3
  2. Access Solution of Introduction to Trigonometry Class 10 Chapter 8 Exercise: 8.3
  3. Topics covered in Chapter 8 Introduction to Trigonometry: Exercise 8.3
  4. NCERT Solutions of Class 10 Subject Wise
  5. NCERT Exemplar Solutions of Class 10 Subject-Wise
NCERT Solutions for Exercise 8.4 Class 10 Maths Chapter 8 - Introduction to Trigonometry
exercise 8.4

Exercise 8.3 in the NCERT Solutions for Class 10 presents three important identities which state \(\sin^2 \theta + \cos^2 \theta = 1\) alongside \(1 + \tan^2 \theta = \sec^2 \theta\) and \(1 + \cot^2 \theta = \csc^2 \theta\). The relationships teach students methods to simplify problems through systematic verification of given expressions. The exercise functions as an essential tool for solidifying the knowledge described in NCERT Books. This exercise represents a fundamental requirement for solving trigonometric problems, which progress into height and distance calculations and advanced mathematical concepts.

NCERT Solutions Class 10 Maths Chapter 8: Exercise 8.3

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Access Solution of Introduction to Trigonometry Class 10 Chapter 8 Exercise: 8.3

Q 1: Express the trigonometric ratios sinA,secA and tanA in terms of cotA .

Answer:

We know that csc2Acot2A=1
(i)
1sin2A=1+cot2Asin2A=11+cot2AsinA=11+cot2A

(ii) We know the identity of
sec2Atan2A=11cos2A=1+tan2A=1+1cot2Acot2A1+cot2A=cos2AcotA1+cot=cosA

(iii) tanA=1cotA

Q 2: Write all the other trigonometric ratios of A in terms of secA .

Answer:

We know that the identity sin2A+cos2=1

sin2A=1cos2sin2A=11sec2A

=sec2A1sec2A

sinA=sec2A1sec2A

=sec2A1secA

cosecA=secAsec2A1

tanA=sinAcosA=sec2A1

cotA=1sec2A1

Q 3: Choose the correct option. Justify your choice (i)9sec2A9tan2A=

(A) 1 (B) 9 (C) 8 (D) 0

Answer:

The correct option is (B) = 9

9sec2A9tan2A=9(sec2Atan2A) .............(i)

And it is known that sec2Atan2A=1

Therefore, equation (i) becomes, 9×1=9

Q 3: Choose the correct option. Justify your choice (ii)(1+tanθ+secθ)(1+cotθcosecθ)=

(A) 0 (B) 1 (C) 2 (D) –1

Answer:

The correct option is (C)

(1+tanθ+secθ)(1+cotθcosecθ) .......................(i)

we can write his above equation as;

=(1+sinθ/cosθ+1/cosθ)(1+cosθ/sinθ1/sinθ)=(1+sinθ+cosθ)cosθ.sinθ×((sinθ+cosθ1sinθ.cosθ)=(sinθ+cosθ)212sinθ.cosθ=sin2θ+cos2θ+2sinθ.cosθ1sinθ.cosθ=2×sinθ.cosθsinθ.cosθ = 2

Q 3: Choose the correct option. Justify your choice (iii)(secA+tanA)(1sinA)=

(A)secA (B)sinA (C)cosecA (D)cosA

Answer:

The correct option is (D)

(secA+tanA)(1sinA)=

(1cosA+sinAcosA)(1sinA)1+sinAcosA(1sinA)1sin2AcosAcosA

Q 3: Choose the correct option. Justify your choice (iv)1+tan2A1+cot2A=

(A)sec2A (B)1 (C)cot2A (D)tan2A

Answer:

The correct option is (D)

1+tan2A1+cot2A ..........................eq (i)

The above equation can be written as;

We know that cotA=1tanA

therefore,

1+tan2A1+1tan2Atan2A×(1+tan2A1+tan2A)tan2A

Q 4: Prove the following identities, where the angles involved are acute angles for which the expressions are defined (i)(cscθcotθ)2=1cosθ1+cosθ

Answer:

We need to prove-

(cscθcotθ)2=1cosθ1+cosθ

Now, taking LHS,

(cscθcotθ)2=(1sinθcosθsinθ)2

=(1cosθsinθ)2=(1cosθ)(1cosθ)sin2θ

=(1cosθ)(1cosθ)1cos2θ=(1cosθ)(1cosθ)(1cosθ)(1+cosθ)=1cosθ1+cosθ

LHS = RHS

Hence proved.

Q 4: Prove the following identities, where the angles involved are acute angles for which the expressions are defined (ii)cosA1+sinA+1+sinAcosA=2secA

Answer:

We need to prove-

cosA1+sinA+1+sinAcosA=2secA

Taking LHS;

=cos2A+1+sin2A+2sinAcosA(1+sinA)=2(1+sinA)cosA(1+sinA)=2/cosA=2secA

= RHS

Hence proved.

Q 4: Prove the following identities, where the angles involved are acute angles for which the expressions are defined (iii)tanθ1cotθ+cotθ1tanθ=1+secθcscθ

[ Hint: Write the expression in terms of sinθ and cosθ ]

Answer:

We need to prove-

tanθ1cotθ+cotθ1tanθ=1+secθcosecθ

Taking LHS;

tan2θtanθ1+1tanθ(1tanθ) tan3θtan4θ+tanθ1(tanθ1).tanθ.(1tanθ)(tan3θ1)(1tanθ)tanθ.(tanθ1)(1tanθ)

By using the identity a 3 - b 3 =(a - b) (a 2 + b 2 +ab)

(tanθ1)(tan2θ+1+tanθ)tanθ(tanθ1a)tanθ+1+1tanθ1+1+tan2θtanθ1+sec2θ×1tanθ1+secθ.cscθ

= RHS

Hence proved.

Q 4: Prove the following identities, where the angles involved are acute angles for which the expressions are defined (iv)1+secAsecA=sin2A1cosA

[ Hint : Simplify LHS and RHS separately]

Answer:

We need to prove-

1+secAsecA=sin2A1cosA

Taking LHS;

1+secAsecA(1+1cosA)/secA1+cosA

Taking RHS;

We know that identity 1cos2θ=sin2θ

sin2A1cosA1cos2A1cosA(1cosA)(1+cosA)(1cosA)1+cosA

LHS = RHS

Hence proved.

Q 4: Prove the following identities, where the angles involved are acute angles for which the expressions are defined (v)cosAsinA+1cosA+sinA1=cscA+cotA , using the identity csc2A=1+cot2A

Answer:

We need to prove -

cosAsinA+1cosA+sinA1=cosecA+cotA

Dividing the numerator and denominator by sinA , we get;

=cotA1+cscAcotA+1cscA=(cotA+cscA)(csc2Acot2A)cotA+1cscA=(cscA+cotA)(1cscA+cotA)cotA+1cscA=cscA+cotA=RHS

Hence Proved.

Q 4: Prove the following identities, where the angles involved are acute angles for which the expressions are defined (vi)1+sinA1sinA=secA+tanA

Answer:

We need to prove -

1+sinA1sinA=secA+tanA

Taking LHS;

By rationalising the denominator, we get;

=1+sinA1sinA×1+sinA1+sinA=(1+sinA)21sin2A=1+sinAcosA=secA+tanA=RHS

Hence proved.

Q 4: Prove the following identities, where the angles involved are acute angles for which the expressions are defined. (vii)sinθ2sin3θ2cos3θcosθ=tanθ

Answer:

We need to prove -

sinθ2sin3θ2cos3θcosθ=tanθ

Taking LHS;

[we know the identity cos2θ=2cos2θ1=cos2θsin2θ ]

sinθ(12sin2θ)cosθ(2cos2θ1)sinθ(sin2θ+cos2θ2sinθ)cosθ.cos2θsinθ.cos2θcosθ.cos2θtanθ=RHS

Hence proved.

Q 4: Prove the following identities, where the angles involved are acute angles for which the expressions are defined (viii)(sinA+cscA)2+(cosA+secA)2=7+tan2A+cot2A.

Answer:

Given the equation,

(sinA+cscA)2+(cosA+secA)2=7+tan2A+cot2A ..................(i)

Taking LHS;

(sinA+cscA)2+(cosA+secA)2

sin2A+csc2A+2+cos2A+sec2A+21+2+2+(1+cot2A)+(1+tan2A)

[since sin2θ+cos2θ=1,csc2θcot2θ=1,sec2θtan2θ=1 ]

7+csc2A+tan2A=RHS

Hence proved

Q 4: Prove the following identities, where the angles involved are acute angles for which the expressions are defined (ix)(cosecAsinA)(secAcosA)=1tanA+cotA

[ Hint : Simplify LHS and RHS separately]

Answer:

We need to prove-

(coescAsinA)(secAcosA)=1tanA+cotA

Taking LHS;

(cosecA1cscA)(secA1secA)(cosec21)cosecA×sec2A1secAcot2AcosecA.tan2AsecAsinA.cosA

Taking RHS;

1sinA/cosA+cosA/sinAsinA.cosAsin2A+cos2AsinA.cosA

LHS = RHS

Hence proved.

Q 4: Prove the following identities, where the angles involved are acute angles for which the expressions are defined (x)(1+tan2A1+cot2A)=(1tanA1cotA)2=tan2A

Answer:

We need to prove,

(1+tan2A1+cot2A)=(1tanA1cotA)2=tan2A

Taking LHS;

1+tan2A1+cot2A=sec2Acsc2A=tan2A

Taking RHS;

=(1tanA1cotA)2=(1sinA/cosA1cosA/sinA)2=(cosAsinA)2(sin2A)(sinAcosA)2(cos2A)=tan2A

LHS = RHS

Hence proved.


Also read-

Topics covered in Chapter 8 Introduction to Trigonometry: Exercise 8.3

1. Understanding basic trigonometric identities: You must learn the essential three trigonometric identities and understand their value for problem simplification and solution.

2. Verifying trigonometric identities: Perform proofs which establish two sides of an equation to be equal by using proper transformations in combination with trigonometric relations.

3. Developing logical and proof-solving skills: Step-by-step logical verification belongs to the set of skills that students need to develop for checking identities and producing mathematical proofs.

4. Preparing for advanced applications: Construct a strong foundation to solve problems utilising height-distance relationships alongside trigonometric equation methods, as well as actual angle measurement scenarios.

Check Out-

NCERT Exemplar Solutions of Class 10 Subject-Wise

Students must check the NCERT Exemplar solutions for class 10 of the Mathematics and Science Subjects.

Frequently Asked Questions (FAQs)

Q: Can I use a calculator for solving trigonometric identities in the exam?
A:

No, calculators are not allowed in Class 10 board exams, so you must rely on your memorised values and identities.

Q: Are the questions in Exercise 8.3 important for board exams?
A:

Yes, Exercise 8.3 is very important as identity-based questions frequently appear in CBSE Class 10 board exams

Q: Do I need to memorize all trigonometric identities?
A:

You should at least remember the fundamental identities and standard angle values to solve Exercise 8.3 effectively.

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